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1 year ago

12

Look at the attachment! This is algebra. 10 points!

1 answer:

boyakko [2]1 year ago

6 0

If x∆y = 3x - y², then

5∆1 = 3×5 - 1² = 15 - 1 = 14

and

14∆6 = 3×14 - 6² = 42 - 36 = 6

So (5∆1)∆6 = 6.

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Chase has been playing a game where he can create towns and help his empire expand. Each town he has allows him to create 1.13 t

riadik2000 [5.3K]

Well the way I see it is he started with 4 villager so I would see the equation to be

1.13x times 17+4 would equal the amount of villagers he could have which would be 23.21 as for the equation it would be 1.13x times 17 plus four equals the number of villagers total

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3 years ago

Look at the picture.....

tresset_1 [31]

Answer:

Step-by-step explanation:

y = a|x-h| + k

(h,k) is the vertex

There's no standard formula for absolute values. I just made it up as an example, pretty much.

Since a is negative, the function opens downward.

h = -2, k = 0, so the vertex is at (-2,0)

5 0

2 years ago

Read 2 more answers

HElppppp pleaseee......

LiRa [457]

Answer: \sqrt[4]{xy^{2} }

Step-by-step explanation: Siplified

6 0

2 years ago

I need some help with this.

Alenkasestr [34]

Answer:

b is 16 d is 13

Step-by-step explanation:

ok is a baby girl that can I be happy

7 0

2 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers